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  • Two moles of helium gas undergo a cyclic process as shown in the figure. Assume the gas to be i

Two moles of helium gas undergo a cyclic process as shown in the figure. Assume the gas to be ideal. Take \mathrm{\mathrm{R}=8.32 \mathrm{Jmol}^{-1}}

                             

.

Calculate the net change in the heat energy in the process

Option: 1

\mathrm{1153.4\: J}


Option: 2

\mathrm{3460.2\: J}


Option: 3

\mathrm{4613.6\: J}


Option: 4

0


Answers (1)

best_answer

Number of moles, \mathrm{n}=2

Helium is a mono-atomic gas, therefore,

Since, \mathrm{\quad \mathrm{C}_{\mathrm{V}}=\frac{3}{2} \mathrm{R} ; \mathrm{C}_{\mathrm{P}}=\frac{5}{2} \mathrm{R}}

The gas undergoes cyclic process.

Since, internal energy is property of the sytem, the net change in internal energy during the cyclic process is zero.

The net change in the heat energy is equal to the net work done.

\mathrm{(\Delta \mathrm{Q})_{\mathrm{Net}} =(\Delta \mathrm{Q})_{\mathrm{AB}}+\left(\Delta \mathrm{Q}_{\mathrm{BC}}\right)+\left(\Delta \mathrm{Q}_{\mathrm{CD}}\right)+\left(\Delta \mathrm{Q}_{\mathrm{DA}}\right) }

\mathrm{(\Delta \mathrm{Q})_{\mathrm{AB}}= \mathrm{n} \times \mathrm{C}_{\mathrm{P}} \times\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right) }

                \mathrm{ =2 \times \frac{5}{2} \times 8.32(400-300)=4160 \mathrm{~J}}

Since Process \mathrm{BC} is isothermal, therefore \Delta \mathrm{U}=0

\mathrm{(\Delta \mathrm{Q})_{\mathrm{BC}}= (\Delta \mathrm{W})_{\mathrm{BC}} }

                \mathrm{ = \mathrm{nRT} \ln \left(\frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)=\mathrm{nRT} \ln \left(\frac{\mathrm{P}_{\mathrm{B}}}{\mathrm{P}_{\mathrm{C}}}\right) }

                \mathrm{ =2 \times 8.32 \times 400 \ln \left(\frac{2}{1}\right)=4613.6 \mathrm{~J} }

\mathrm{ \left(\Delta \mathrm{Q}_{\mathrm{CD}}\right) =\mathrm{nC}_{\mathrm{p}}\left(\mathrm{T}_{\mathrm{D}}-\mathrm{T}_{\mathrm{C}}\right) }

                \mathrm{ =2 \times \frac{5}{2} \times 8.32 \times(300-400)=-4160 \mathrm{~J} }

\mathrm{ (\Delta Q)_{D A}= n R T \ln \left(\frac{\mathrm{P}_{\mathrm{D}}}{\mathrm{P}_{\mathrm{A}}}\right) }

                 \mathrm{ =2 \times 8.32 \times 300 \ln \left(\frac{2}{1}\right)=-3460.2 \mathrm{~J} }

\mathrm{(\Delta \mathrm{W})_{\text {Net }}= 4160+4613.6-4160-3460.2 }

                    \mathrm{ =1153.4 \mathrm{~J}}






 






 

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